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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 361-390 of 483 questions

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361 Medium
361 of 483
Analyzing +4 | -1
The bob of a pendulum is in the lowest position and has a horizontal velocity of gL\sqrt {gL} where LL is length of the pendulum and gg is acceleration due to gravity is given to it. Then the bob
View answer and explanation
Correct option: C
Oscillates about the lowest position as mean position
Option C (Oscillates about the lowest position as mean position) is correct because the initial velocity is insufficient to complete a full circle, resulting in oscillation.
362 Medium
362 of 483
Application +4 | -1
A bob of mass m suspended by a thread of length ll undergoes simple harmonic oscillations with time period TT. If the bob is immersed in a liquid that has density 1/41/4 times that of the bob and the length of the thread is increased by 1/3  rd1\mathrm{/}3\,\,\mathrm{rd} of the original length, then the time period of the simple harmonic oscillations will be
View answer and explanation
Correct option: D
43T\frac{4}{3}T
Option D (43T\frac{4}{3}T) is correct because the time period of a simple pendulum changes due to the effective gravity in the liquid and the change in length.
363 Medium
363 of 483
Application +4 | -1
A particle is executing simple harmonic motion with frequency ff . The frequency at which its kinetic energy change into potential energy is
View answer and explanation
Correct option: C
2  f2\,\,f
Option C (2  f2\,\,f) is correct because the frequency of kinetic and potential energy interchange in SHM is twice the frequency of the oscillating particle.
364 Medium
364 of 483
Understanding +4 | -1
If a simple harmonic is represented by d2xdt2\frac{{{d^2}x}}{{d{t^2}}}+αx=0+ \alpha x = 0,its time period is
View answer and explanation
Correct option: B
2πα\frac{{2\pi }}{{\sqrt \alpha }}
Option B is correct because the time period of a simple harmonic motion is given by T=2πωT = \frac{{2\pi}}{{\omega}}, where ω=α\omega = \sqrt{{\alpha}} in this case.
365 Medium
365 of 483
Application +4 | -1
Two pendulums of length 121 cm and 100 cm start vibrating. At same instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum, the two will be in phase in the mean position?
View answer and explanation
Correct option: B
11
Option B (11) is correct because it satisfies the condition where the shorter pendulum completes an integer number of vibrations while the longer pendulum completes one less to be in phase again.
366 Medium
366 of 483
Remembering +4 | -1
If a body oscillates at the angular frequency ωd{\omega _d} of the driving force, then the oscillations are called
View answer and explanation
Correct option: C
Forced oscillations
Option C (Forced oscillations) is correct because forced oscillations occur when a body oscillates at the driving force's frequency.
367 Medium
367 of 483
Understanding +4 | -1
Which one of the following is simple harmonic?
View answer and explanation
Correct option: D
Motion of oscillating liquid column in U-tube
Option D (Motion of oscillating liquid column in U-tube) is correct because the restoring force is proportional to the displacement from the equilibrium position.
368 Medium
368 of 483
Understanding +4 | -1
Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion
View answer and explanation
Correct option: D
When v is maximum, a is zero
Option D (When v is maximum, a is zero) is correct because in SHM, velocity is maximum at the equilibrium position where acceleration is zero.
369 Medium
369 of 483
Application +4 | -1
The time period of a simple pendulum measured inside a stationary elevator is 4 s4 {~s}. If the lift start accelerating downwards with an acceleration 3g/43 g / 4, the time period will be
View answer and explanation
Correct option: B
8 s8 {~s}
Option B (8 s) is correct because the effective acceleration changes when the elevator accelerates downwards, affecting the time period of the pendulum.
370 Easy
370 of 483
Analyzing +4 | -1
Which of the following quantities is always negative in SHM?
Here, s\mathbf{s} is displacement from mean position.
View answer and explanation
Correct option: C
a⋅s\mathbf{a} \cdot \mathbf{s}
Option C (a⋅s\mathbf{{a}} \cdot \mathbf{{s}}) is correct because in SHM, acceleration and displacement are always in opposite directions, resulting in a negative dot product.
371 Medium
371 of 483
Analyzing +4 | -1
Assertion: A heavy metal sphere is suspended by a spring. When pulled down a little and released, it oscillates up and down with a frequency f.f. If it is taken to the moon where the acceleration due to gravity is g/6,g/6, the frequency of oscillation is 6f\sqrt{6}f.
Reason: The frequency of oscillation of the spring depends only on the mass of the sphere and spring constant.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because the frequency of oscillation does not depend on gravity, and the frequency would remain the same on the moon.
372 Medium
372 of 483
Analyzing +4 | -1
The time period of a simple pendulum is TT. If its point of suspension is moved upward according to relation y=λt2y=\lambda t^2 where λ\lambda is a constant then its new time period T1T_1.
View answer and explanation
Correct option: C
is less than T{T}
Option C is correct because the effective acceleration due to gravity increases when the suspension point moves upwards, leading to a decrease in the time period of the pendulum.
373 Medium
373 of 483
Analyzing +4 | -1
A simple pendulum is made of bob which is a hollow sphere full of sand suspended by means of a wire. If all the sand is drained out, the period of the pendulum will
View answer and explanation
Correct option: C
Remain same
Option C (Remain same) is correct because the length of the pendulum, measured from the pivot to the center of mass, does not change when the sand is drained.
374 Medium
374 of 483
Application +4 | -1
A mass of 2 kg attached to a spring of force constant 260∘   Nm−1260^\circ \,\,\,N{m^{ - 1}} makes 200 oscillations. What is the time taken [in seconds] nearly?
View answer and explanation
Correct option: D
110
Option D (110) is correct because the time taken for 200 oscillations is calculated using the formula for the time period of a spring-mass system.
375 Medium
375 of 483
Understanding +4 | -1
A plate oscillates with time period ′T′'T'. Suddenly, another plate put on the first time, then time period
View answer and explanation
Correct option: C
Will be same
Option B (Will increase) is incorrect and option B (Will increase) is correct because adding another plate increases the mass, which affects the time period of oscillation.
376 Medium
376 of 483
Application +4 | -1
Two particles PP and QQ describe SHM of same amplitude aa and frequency vv along the same straight line. The maximum distance between two particle is 2a\sqrt{2} a. The initial phase difference between the particles is
View answer and explanation
Correct option: B
π/2\pi / 2
Option B (π/2\pi / 2) is correct because the maximum distance between the particles is related to the phase difference by 2a=a2(1−cos⁡ϕ)\sqrt{2}a = a\sqrt{2(1-\cos{\phi})}.
377 Medium
377 of 483
Application +4 | -1
For a SHO, if the magnitudes of maximum velocity and acceleration are equal, its frequency of oscillations is
View answer and explanation
Correct option: B
12π  Hz\frac{1}{{2\pi }}\,\,{\text{Hz}}
Option B (12π Hz\frac{{1}}{{2\pi }}\,{\text{{Hz}}}) is correct because equating the magnitudes of maximum velocity and acceleration in SHM leads to ω=1\omega = 1, and then using ω=2πf\omega = 2\pi f gives the frequency.
378 Medium
378 of 483
Understanding +4 | -1
The elongation of spring of length LL and of negligible mass due to a force is xx. the spring is cut into two pieces of length in ratio 1:n1: n. The ratio of the respective spring constants is:
View answer and explanation
Correct option: A
n:1{n}: 1
Option A is correct because the spring constant is inversely proportional to the length of the spring, so when the spring is cut into a ratio of 1:n, the spring constants are in the ratio of n:1.
379 Medium
379 of 483
Application +4 | -1
A particle executes SHM with a time period of 2 s2 \mathrm{~s} and amplitude 5 cm.5 \mathrm{~cm}. Find displacement after 1/31 / 3 second; starting from the mean position.
View answer and explanation
Correct option: A
4.33 cm4.33 \mathrm{~cm}
Option A (4.33 cm4.33 \mathrm{~cm}) is correct because it is the displacement calculated using the SHM equation x=Asin⁡ωtx=A \sin \omega t with the given time period, amplitude, and time.
380 Medium
380 of 483
Application +4 | -1
F−xF-x equation of a body in SHM along x{x}-axis is F+12x=0F+12 x=0. Here, FF is in newton and xx in metre. If the mass of the body is 3 kg3 {~kg}, then the time period of oscillations is;
View answer and explanation
Correct option: C
πs\pi {s}
Option C (π\pi s) is correct because it is calculated using the formula for the time period of SHM, T=2πmKT = 2\pi \sqrt{{\frac{{m}}{{K}}}} after finding K from the given equation.
381 Medium
381 of 483
Application +4 | -1
The force constant of two springs are  K1\;{K_1} and K2{K_2} Both are stretched till their elastic energies are equal. If the stretching forces are F1{F_1} and F2,{F_2}, then F1:F2{F_1}:{F_2} is
View answer and explanation
Correct option: C
K1:K2\sqrt {{K_1}} :\sqrt {{K_2}}
Option C (K1:K2\sqrt {{K_1}} :\sqrt {{K_2}}) is correct because it is derived from equating the elastic potential energies and relating them to the stretching forces.
382 Easy
382 of 483
Analyzing +4 | -1
A small ball is dropped from a certain height on the surface of a non-viscous liquid of density less than the density of ball. The motion of the ball is
View answer and explanation
Correct option: B
periodic but not SHM
Option B (periodic but not SHM) is correct because after attaining terminal velocity, the ball moves with constant velocity, resulting in periodic but non-SHM motion.
383 Medium
383 of 483
Application +4 | -1
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement
View answer and explanation
Correct option: C
a2\frac{a}{{\sqrt 2 }}
Option C (a2\frac{a}{{\sqrt 2 }}) is correct because at this displacement, the potential energy and kinetic energy of the particle in SHM are equal.
384 Medium
384 of 483
Understanding +4 | -1
What is the phase difference between velocity and acceleration of a particle executing SHM?
View answer and explanation
Correct option: C
π/2\pi / 2
Option C is correct because in SHM, velocity leads displacement by π/2\pi/2 and acceleration leads velocity by π/2\pi/2, hence the phase difference between velocity and acceleration is π/2\pi / 2.
385 Easy
385 of 483
Understanding +4 | -1
A horizontally placed spring mass system has time period TT. The same system is now placed on a car moving with acceleration aa in horizontal direction. Then,
View answer and explanation
Correct option: C
time period will remain constant
Option C (time period will remain constant) is correct because the time period of a spring-mass system depends only on the mass and spring constant, not on external acceleration.
386 Medium
386 of 483
Understanding +4 | -1
A body is executing simple harmonic motion with frequency 'nn' the frequency of its potential energy is
View answer and explanation
Correct option: D
2n2n
Option D (2n2n) is correct because the frequency of potential energy in SHM is twice the frequency of the displacement.
387 Medium
387 of 483
Application +4 | -1
A particle performs Simple Harmonic Motion of amplitude A{A} and angular frequency ω\omega along a straight line. When it is at a distance 32 A\frac{\sqrt{3}}{2} {~A} from mean position, its kinetic energy gets increased suddenly by an amount 12mω2A2\frac{1}{2} m \omega^2 A^2, due to an impulsive force. Then new amplitude of oscillations of the particle will be :
View answer and explanation
Correct option: C
2 A\sqrt{2} {~A}
Option C (2 A\sqrt{2} {~A}) is correct because the new amplitude is calculated by equating the new total energy after the impulse to the potential energy at the new amplitude.
388 Medium
388 of 483
Application +4 | -1
A second's pendulum is taken in a carriage. Find the period of oscillation when the carriage moves with an acceleration of 4 ms−24 \mathrm{~ms}^{-2} (i) vertically upwards (ii) vertically downwards, and (iii) in a horizontal direction.
View answer and explanation
Correct option: A
1.69 s;2.59 s;1.92 s1.69 \mathrm{~s}; 2.59 \mathrm{~s}; 1.92 \mathrm{~s}
Option A (1.69 s;2.59 s;1.92 s1.69 \mathrm{~s}; 2.59 \mathrm{~s}; 1.92 \mathrm{~s}) is correct because it correctly calculates the period of oscillation of a second's pendulum in a carriage moving with different accelerations.
389 Medium
389 of 483
Remembering +4 | -1
The differential equation of a particle executing simple harmonic motion along yy-axis is
View answer and explanation
Correct option: A
d2ydt2+ω2y=0\frac{d^2 y}{d t^2}+\omega^2 y=0
Option A (d2ydt2+ω2y=0\frac{{d^2 y}}{{d t^2}}+\omega^2 y=0) is the correct differential equation for a particle executing simple harmonic motion along the y-axis because it represents the relationship between acceleration and displacement in SHM.
390 Medium
390 of 483
Remembering +4 | -1
Spring force is a
View answer and explanation
Correct option: B
Conservative force
Option B (Conservative force) is correct because spring force is a conservative force, meaning the work done by it is path-independent and can be expressed as a potential energy.
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